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This deck focuses on Direction Of Reversible Reactions, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.
Study Direction Of Reversible Reactions in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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What happens if a catalyst is added to a reaction at equilibrium?
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No effect on the equilibrium position. Equilibrium position remains unchanged with catalyst addition.
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This deck focuses on Direction Of Reversible Reactions, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: No effect on the equilibrium position. Equilibrium position remains unchanged with catalyst addition.
Answer: To predict the direction of a reaction's shift to reach equilibrium. Compares current concentrations to equilibrium to determine reaction direction.
Answer: No effect on the equilibrium position. Pure liquids have constant activity and don't appear in Kc.
Answer: Shifts toward the side with fewer moles of gas. Higher pressure favors the side with fewer gas molecules.
Answer: Kp=Kc(RT)Δn. Standard relationship connecting concentration and pressure equilibrium constants.
Answer: Kp=Kc(RT)Δn. Conversion formula where R is gas constant and T is temperature.
Answer: Kc increases. Higher temperature favors the endothermic direction.
Answer: Equilibrium constant in terms of partial pressures. Uses partial pressures instead of molar concentrations.
Answer: Kc decreases. Higher temperature shifts equilibrium away from the exothermic direction.
Answer: Shifts equilibrium to the right. System minimizes pressure by favoring the side with fewer gas molecules.
Answer: Shifts equilibrium to the right. System reduces pressure by favoring the side with more gas molecules.
Answer: It does not affect the position of equilibrium. Catalyst speeds both forward and reverse reactions equally.
Answer: The equilibrium will shift to the right. System responds by producing more of the removed product.
Answer: The equilibrium favors products. Large Kc indicates products are thermodynamically favored.
Answer: The reaction is reversible. Double arrows indicate the reaction can proceed in both directions.
Answer: Kc=[A]a[B]b[C]c[D]d. Products raised to stoichiometric powers divided by reactants raised to powers.
Answer: Increases the rate but does not change the position. Catalyst accelerates approach to equilibrium without changing final position.
Answer: The equilibrium favors reactants. Small Kc indicates reactants are thermodynamically favored.
Answer: An equilibrium will shift to oppose changes in concentration, temperature, or pressure. System adjusts to counteract any imposed change or stress.
Answer: Shifts toward the side with fewer moles of gas. Compression increases pressure, favoring the side with fewer gas moles.
Answer: Unitless. Concentration units cancel out in the equilibrium expression.
Answer: Shifts equilibrium to the right. Heat is a reactant, so adding heat favors product formation.
Answer: Kc increases. Higher temperature favors the endothermic direction.
Answer: The equilibrium will shift to the left. System responds by consuming more products to replace removed reactant.
Answer: Change in moles of gas: (moles of gaseous products) - (moles of gaseous reactants). Difference in moles of gaseous species between products and reactants.
Answer: A reversible reaction. Double arrows indicate the reaction proceeds in both directions.
Answer: Rates of forward and reverse reactions are equal. Forward and reverse reactions occur at the same rate continuously.
Answer: Kc increases. Lower temperature favors the exothermic direction.
Answer: No effect on the equilibrium position. Pure solids have constant activity and don't appear in Kc.
Answer: Shifts equilibrium to the right. System consumes added reactant by forming more products.
Answer: The reaction will shift to the left. More reactants need to form to reach equilibrium.
Answer: No effect on the equilibrium position. Inert gas doesn't change partial pressures of reactants or products.
Answer: It does not change the value of Kc. Kc depends only on temperature, not concentration changes.
Answer: Shifts equilibrium to the left. System consumes excess product by favoring the reverse reaction.
Answer: Shifts toward the side with fewer moles of gas. Compression increases pressure, favoring the side with fewer gas moles.
Answer: The reaction is at equilibrium. Forward and reverse reaction rates are equal, no net change.
Answer: Temperature. Only temperature changes affect the equilibrium constant value.
Answer: Shifts equilibrium to the right. Heat is a product, so removing heat favors product formation.
Answer: The reaction will shift to the right. More products need to form to reach equilibrium.
Answer: Shifts toward the side with fewer moles of gas. Reduced volume increases pressure, favoring fewer gas molecules.